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Question Number 40709 by math khazana by abdo last updated on 26/Jul/18

$$calculate{lim}_{x\to 0}\frac{cos(x-sinx)-1}{x^2}$$

Commented by maxmathsup by imad last updated on 26/Jul/18

$$wehavesinx=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)!}{x}^{2n+1}\Rightarrow sinx=x-\frac{1}{3!}{x}^3+o\left(x^5\right)\Rightarrow$$$$x-sinx=\frac{x^3}{6}+o\left(x^5\right)\Rightarrow cos(x-sinx)=cos(\frac{x^3}{6}+o\left(x^5\right))\sim 1-\frac{x^6}{12}(x\to 0)\Rightarrow$$$$cos(x-sinx)-1\sim -\frac{x^6}{12}\Rightarrow \frac{cos(x-sinx)-1}{x^2}\sim \frac{x^4}{12}\Rightarrow$$$${lim}_{x\to 0}\frac{cos(x-sinx)-1}{x^2}=0$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jul/18

$$\underset{x\to 0}{\lim }\frac{-2{sin}^2\left(\frac{x-sinx}{2}\right)}{{\left(\frac{x-sinx}{2}\right)}^2}\times \frac{{\left(\frac{x-sinx}{2}\right)}^2}{x^2}$$$$whenx\to 0\frac{x-sinx}{2}\to 0$$$$t=\frac{x-sinx}{2}$$$$\underset{t\to 0}{\lim }\frac{-2{sin}^{2}t}{t^2}\times \underset{x\to 0}{\lim }\frac{x^2{(1-\frac{sinx}{x})}^2}{4x^2}$$$$=-2\times \frac{1}{4}\times (1-1)=0$$

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